中文

An efficient algorithm for the computation of Bernoulli numbers

数论 2007-05-23 v2 经典分析与常微分方程

摘要

This article gives a direct formula for the computation of B(n) using the asymptotic formula B(n)2n!πn2nB (n) \approx 2 {\frac {n!}{{\pi}^{n}{2}^{n}}} where n is even and n>>1n >> 1. This is simply based on the fact that ζ(n)\zeta (n) is very near 1 when n is large and since B(n)=2ζ(n)n!πn2nB (n) = 2 {\frac {\zeta (n) n!}{{\pi}^{n}{2}^{n}}} exactly. The formula chosen for the Zeta function is the one with prime numbers from the well-known Euler product for ζ(n)\zeta (n). This algorithm is far better than the recurrence formula for the Bernoulli numbers even if each B(n) is computed individually. The author could compute B(750,000)B (750,000) in a few hours. The current record of computation is now (as of Feb. 2007) B(5,000,000)B (5,000,000) a number of (the numerator) of 27332507 decimal digits is also based on that idea.

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引用

@article{arxiv.math/0702300,
  title  = {An efficient algorithm for the computation of Bernoulli numbers},
  author = {Greg Fee and Simon Plouffe},
  journal= {arXiv preprint arXiv:math/0702300},
  year   = {2007}
}